A131293 -id:A131293 - cp's OEIS frontend

A063896 a(n) = 2^Fibonacci(n) - 1.

0, 1, 1, 3, 7, 31, 255, 8191, 2097151, 17179869183, 36028797018963967, 618970019642690137449562111, 22300745198530623141535718272648361505980415

Offset: 0

Robert G. Wilson v, Aug 29 2001

The recurrence can also be written a(n)+1 = (a(n-1)+1)*(a(n-2)+1) or log_p(a(n)+1) = log_p(a(n-1)+1) + log_p(a(n-2)+1), respectively. Setting a(1)=p-1 for any natural p>1, it follows that log_p(a(n)+1)=Fibonacci(n). Hence any other sequence p^Fibonacci(n)-1 could also serve as a valid solution to that recurrence, depending only on the value of the term a(1). - Hieronymus Fischer, Jun 27 2007
Written in binary, a(n) contains Fibonacci(n) 1's. Thus the sequence converted to base-2 is A007088(a(n)) = 0, 1, 1, 11, 111, 11111, 11111111, ... . - Hieronymus Fischer, Jun 27 2007
In general, if b(n) is defined recursively by b(0) = p, b(1) = q, b(n) = b(n-1)*b(n-2) + b(n-1) + b(n-2) for n >= 2 then b(n) = p^Fibonacci(n-1) * q^Fibonacci(n) - 1. - Rahul Goswami, Apr 15 2020
a(n) is also the numerator of the continued fraction [2^F(0), 2^F(1), 2^F(2), 2^F(3), ..., 2^F(n-2)] for n>0. For the denominator, see A005203. - Chinmay Dandekar and Greg Dresden, Sep 19 2020

Alois P. Heinz, Table of n, a(n) for n = 0..18
M. Tamba and Y. S. Valaulikar, A nonlinear extension of Fibonacci sequence, Turkish Journal of Analysis and Number Theory, Vol. 4, No. 4, 2016.

Cf. A000045 (Fibonacci), A000225, A000301, A005203, A061107.

See A131293 for a base-10 analog with Fib(n) 1's.

a:= n-> 2^(<<0|1>, <1|1>>^n)[1,2]-1:
seq(a(n), n=0..15);  # Alois P. Heinz, Aug 12 2017

2^Fibonacci[Range[0,15]]-1 (* Harvey P. Dale, May 20 2014 *)
RecurrenceTable[{a[0] == 0, a[1] == 1, a[n] == (a[n - 1] + 1)*(a[n - 2] + 1) - 1}, a[n], {n, 0, 12}] (* Ray Chandler, Jul 30 2015 *)

a(n) = 2^fibonacci(n) - 1 \\ Charles R Greathouse IV, Oct 03 2016

The solution to the recurrence a(0) = 0; a(1) = 1; a(n) = a(n-1)*a(n-2) + a(n-1) + a(n-2).
a(n) = A000301(n) - 1. - R. J. Mathar, Apr 26 2007
a(n) = a(n-2)*2^ceiling(log_2(a(n-1))) + a(n-1) for n>1. - Hieronymus Fischer, Jun 27 2007
a(n) = A000225(A000045(n)). - Alois P. Heinz, Mar 19 2020

A108047 Concatenation of the previous pair of numbers, with the first two terms both 1.

Original entry on oeis.org

1, 1, 11, 111, 11111, 11111111, 1111111111111, 111111111111111111111, 1111111111111111111111111111111111, 1111111111111111111111111111111111111111111111111111111, 11111111111111111111111111111111111111111111111111111111111111111111111111111111111111111

Offset: 1

Parthasarathy Nambi, Jun 01 2005

The Fibonacci numbers, A000045, represented in base 1 (see A000042).

			The third term is 11 which is the concatenation of the first two terms 1 and 1.

Column b=1 of A214326.
Column k=1 of A214679.
Cf. A037842, A131293.

a(n) = (10^A000045(n)-1)/9.

a(n) = A000042(A000045(n)).

Edited by Jason Kimberley, Dec 15 2012

A037842 Fibonacci numbers in base 1.

Original entry on oeis.org

1, 11, 111, 11111, 11111111, 1111111111111, 111111111111111111111, 1111111111111111111111111111111111, 1111111111111111111111111111111111111111111111111111111

Offset: 0

N. J. A. Sloane, Feb 27 2000

Cf. A108047, A131293.

FromDigits[Table[1,{#}]]&/@Fibonacci[Range[2,10]] (* Harvey P. Dale, Apr 06 2013 *)

a(n+2) = concat( a(n+1), a(n) ), a(0) = 1, a(1) = 11. - corrected by Jason Kimberley, Dec 17 2012

More terms from James Sellers, Mar 01 2000

A214326 Square array read by antidiagonals in which T(n,b) gives the n-th Fibonacci number written in base b with n,b >= 1.

Original entry on oeis.org

1, 1, 1, 1, 1, 11, 1, 1, 10, 111, 1, 1, 2, 11, 11111, 1, 1, 2, 10, 101, 11111111, 1, 1, 2, 3, 12, 1000, 1111111111111, 1, 1, 2, 3, 11, 22, 1101, 111111111111111111111, 1, 1, 2, 3, 10, 20, 111, 10101, 1111111111111111111111111111111111, 1, 1, 2, 3, 5, 13, 31, 210, 100010

Offset: 1

Alois P. Heinz, Jul 24 2012

For b > 10, some terms cannot be properly notated using only decimal characters.

			Square array A(n,b) begins:
              1,    1,   1,  1,  1,  1,  1,  1,  1,  1,  1,  1, ...
              1,    1,   1,  1,  1,  1,  1,  1,  1,  1,  1,  1, ...
             11,   10,   2,  2,  2,  2,  2,  2,  2,  2,  2,  2, ...
            111,   11,  10,  3,  3,  3,  3,  3,  3,  3,  3,  3, ...
          11111,  101,  12, 11, 10,  5,  5,  5,  5,  5,  5,  5, ...
       11111111, 1000,  22, 20, 13, 12, 11, 10,  8,  8,  8,  8, ...
  1111111111111, 1101, 111, 31, 23, 21, 16, 15, 14, 13, 12, 11, ...

Alois P. Heinz, Antidiagonals n = 1..13

Columns b=1-10, 12, 13 give: A108047, A004685, A004686, A004687, A004688, A004689, A004690, A004691, A004692, A000045, A004693, A004694.

Cf. A037842, A131293.

A:= proc(n, b) local f, l; f:= combinat[fibonacci](n);
      if b=1 then parse(cat(1$f))
    else l:= NULL;
         while f>0 do l:= irem(f, b, 'f'), l od;
         parse(cat(l))
      fi
    end:
seq(seq(A(n, 1+d-n), n=1..d), d=1..10);

A334025 a(0)=0, a(1)=1; and a(n) = {2a(n-2), 2a(n-1)}, where {x,y} is the concatenation of x and y.

Original entry on oeis.org

0, 1, 2, 24, 448, 48896, 89697792, 97792179395584, 179395584195584358791168, 195584358791168358791168391168717582336, 358791168391168717582336391168717582336717582336782337435164672

Offset: 0

Jamie Robert Creasey, Apr 14 2020

This sequence, due to the process of concatenating one number with another, bears similarities to A131293 and other familiar sequences. However, unlike A131293, this sequence increases at a faster rate. It happens due to the multiplier applied to the existing terms, which increases the number of digits present in the successive term drastically (see a(7) and a(8)). a(11) is too large to include here and has 102 digits.

			a(2) = {2*a(2-2), 2*a(2-1)} = {2*0, 2*1} = 02 = 2.
a(5) = {2*a(5-2), 2*a(5-1)} = {2*24, 2*448} = 48896.

Cf. A002605, A061107, A108047, A131293, A008352, A113526, A104458.

a[0] = 0; a[1] = 1; a[n_] := a[n] = FromDigits @ Join[IntegerDigits[2*a[n - 2]], IntegerDigits[2*a[n - 1]]]; Array[a, 11, 0] (* Amiram Eldar, Apr 18 2020 *)

A339713 a(n) = (a(n-2) concatenate a(n-1)) for n > 2, with a(1)=1, a(2)=10.

Original entry on oeis.org

1, 10, 110, 10110, 11010110, 1011011010110, 110101101011011010110, 1011011010110110101101011011010110, 1101011010110110101101011011010110110101101011011010110, 10110110101101101011010110110101101101011010110110101101011011010110110101101011011010110

Offset: 1

Wesley Ivan Hurt, Apr 24 2021

Number of digits in a(n) = A000045(n+1). - Michael S. Branicky, Apr 24 2021

a(n) and a(n+1) contain Fibonacci(n) 1's and Fibonacci(n) 0's respectively.

Cf. A000045, A111061 (in decimal), A061107, A131293.

def aupton(terms):
  alst = [1, 10]
  for n in range(3, terms+1): alst.append(int(str(alst[-2])+str(alst[-1])))
  return alst[:terms]
print(aupton(10)) # Michael S. Branicky, Apr 24 2021

cp's OEIS Frontend

A063896 a(n) = 2^Fibonacci(n) - 1.

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A108047 Concatenation of the previous pair of numbers, with the first two terms both 1.

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A037842 Fibonacci numbers in base 1.

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A214326 Square array read by antidiagonals in which T(n,b) gives the n-th Fibonacci number written in base b with n,b >= 1.

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A334025 a(0)=0, a(1)=1; and a(n) = {2*a(n-2), 2*a(n-1)}, where {x,y} is the concatenation of x and y.

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Mathematica

A339713 a(n) = (a(n-2) concatenate a(n-1)) for n > 2, with a(1)=1, a(2)=10.

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Python

A334025 a(0)=0, a(1)=1; and a(n) = {2a(n-2), 2a(n-1)}, where {x,y} is the concatenation of x and y.